Question

a sample of liquid water at 45.0°c is heated to a gas at 150.°c at a constant pressure of 1 atm. create the equation for the total heat required for this process by dragging the equations below into the boxes in the order in which the heat will be used by the sample. it is not necessary to drag into all of the boxes. cₛ(ice) = 2.08 j/mol·k cₛ(liq) = 4.18 j/mol·k cₛ(gas) = 2.11 j/mol·k δh_fus = 6.06 kj/mole δh_vap = 40.7 kj/mole q = heat for the first process + heat for the second process + heat for the third process + heat for the fourth process + heat for the fifth process 2.11 j/mol·k × mass × 50 k moles × 6060 j/mole 2.08 j/mol·k × mass × 45 k 4.18 j/mol·k × mass × 55 k moles × 40700. j/mole 0.5000 reset submit

Explanation:

Step1: Identify process 1: heat liquid water from 45.0°C to 100.0°C

Temperature change: $\Delta T = 100.0^\circ\text{C} - 45.0^\circ\text{C} = 55\ \text{K}$
Heat expression: $4.18\ \text{J/mol·K} \times \text{mass} \times 55\ \text{K}$

Step2: Identify process 2: vaporize liquid water to gas

Enthalpy of vaporization: $\Delta H_{vap} = 40.7\ \text{kJ/mol} = 40700\ \text{J/mol}$
Heat expression: $\text{moles} \times 40700\ \text{J/mole}$

Step3: Identify process 3: heat gas from 100.0°C to 150.0°C

Temperature change: $\Delta T = 150.0^\circ\text{C} - 100.0^\circ\text{C} = 50\ \text{K}$
Heat expression: $2.11\ \text{J/mol·K} \times \text{mass} \times 50\ \text{K}$

Answer:

$q = 4.18\ \text{J/mol·K} \times \text{mass} \times 55\ \text{K} + \text{moles} \times 40700\ \text{J/mole} + 2.11\ \text{J/mol·K} \times \text{mass} \times 50\ \text{K}$

(Note: The initial water is liquid at 45.0°C, so processes involving ice heating or melting are not needed for this scenario.)